Question

The pOH of an aqueous solution of 0.468 M acetylsalicylic acid (aspirin), HC9H7O4, is

please show work

Answer 1

Ka of acetylsalycilic acid = 3.3 x 10^-4

HA <====> H+ + A-

for weak acid Ka = [H+][A-] / [HA]

[H+] = [A-]   ( after dissociation of small amount of weak acid concentration of H+ ion and concentration of conjucate base anion in the solution will be equal)

So Ka = [H+]²/[HA]

[H+]² = Ka*[HA]

[HA] = 0.468 M

[H+]² = 3.3 x 10^-4 * 0.468

[H+] = {3.3 x 10^-4 * 0.468}^1/2

[H+] = 0.0124 M

pH = - log (0.0124)

pH = 1.90

pOH = 14 - pH

pOH = 14 - 1.90

pOH = 12.10